take a ,b to be positive integers (including 0)
i was trying to prove (a^2+b^2)/(ab+1)=integer implies (a^2+b^2)/(ab+1)=square integer
ive come up with two general cases for solutions to form an integer
a=k,b=k^3
AND
a=k^3,b=k^5k
subbing both these cases clearly makes squares i feel I'm close but i just need a bit of hint in proving these are the only solution forms (I'm pretty sure they are though i may be wrong)
number theory hint (no solution wanted)
Announcements  Posted on  

TSR's new app is coming! Sign up here to try it first >>  17102016 

 Follow
 1
 4 weeks ago 4w ago

 Follow
 2
 4 weeks ago 4w ago
Post rating:2 
 Follow
 3
 4 weeks ago 4w ago
(Original post by RDKGames)

 Follow
 4
 3 weeks ago 3w ago
(Original post by chemphys)
I'm reluctant to watch cos it might reveal too much at once lol
Smallish hintSuppose (a^2+b^2) = k(ab+1) (*) for some integer k. Your general approach will be to show that you can find a "smaller" integer solution (A, B) to (*) unless (A, B) "obviously" require that k must be a square.
(Words in quotes are intentionally vague so to spoil as little as possible).Post rating:3 
 Follow
 5
 3 weeks ago 3w ago
(Original post by chemphys)
i feel I'm close but i just need a bit of hint in proving these are the only solution forms (I'm pretty sure they are though i may be wrong)Spoiler:Showa=30, b=112 is a solution that doesn't match either of your forms. 
 Follow
 6
 3 weeks ago 3w ago
(Original post by DFranklin)
Spoiler:Showa=30, b=112 is a solution that doesn't match either of your forms.
oh wait thats just 'one iteration' further in the method i used to find the 2nd form (sum of roots in quadratics)Last edited by chemphys; 3 weeks ago at 17:53. 
 Follow
 7
 3 weeks ago 3w ago
oh i think i might have been on the right lines before i stupidly didn't continue on from the first 2 forms
yeh I've got a sequence of solutions that get 'larger' so i guess the reverse gets a sequence that gets smaller
and theres only so far you can go before u start going into negatives(i.e. 0,0)  i haven't got a formal proof but thats right line?Last edited by chemphys; 3 weeks ago at 17:56. 
 Follow
 8
 3 weeks ago 3w ago
Yes, you should be able to adapt your approach to go down  I don't think you'll get down to (0,0), but you should get to where it's obvious k must be square.

 Follow
 9
 3 weeks ago 3w ago
(Original post by DFranklin)
Yes, you should be able to adapt your approach to go down  I don't think you'll get down to (0,0), but you should get to where it's obvious k must be square.
let a,b satisfy (a^2+b^2)/(ab+1)=k (*) where a>b (only 1 solution where both the same) for now, let k be a fixed positive integer
then a^2+b^2=k(ab+1) (**)
then there exists a pair a',b' satisfying a'^2+b'^2=k(a'b'+1),where b'=b and a'=kba
from (**)/a we get a+b^2/a=kb+k/a so a'=(b^2k)/a<(b^2k)/b<b since k>0
so clearly a' is less than b.
clearly we can keep iterating to get smaller and smaller values for the lower integer of the pair of solutions. eventually we should end up with one of the lower value being the square root of k . if this didn't happen eventually we would get a value that is negative for one of the pair but if one of the pair is negative k can't be positive ,a contradiction since we fixed k to be positive integer .
However, if we get to a stage where the lower value of a pair squared =k then we we get the next iteration to be 0. now this leads to the next lowest number being negative but this still allows k to be positive so no 'breaking' the equation there (I'm kind of artificially allowing one of the terms a, b to be negative for this last step but this is ok because its an exception where k is still positive)
so the only way for k to be positive which it must be is for it to equal some b^2Last edited by chemphys; 3 weeks ago at 16:18. 
 Follow
 10
 3 weeks ago 3w ago
(Original post by chemphys)
..
Some thoughts on things you might want to do to clean things up.
If you want to argue that "we can keep iterating", you need to make it very clear that the original conditions are still satisfied. It's kind of trivial, but making a' = b and b' = kba would mean that your argument showed that a' > b' and so we have a more direct parallel with the initial scenario with a > b.
Also, with a "we can keep iterating", you generally want to pay more attention to "when do/should we stop?" In this case you should find your argument for finding b' fails when b = 0. So you can see that this is when the iteration stops and then you have k = a^2, so k is a square.Last edited by DFranklin; 3 weeks ago at 21:54. 
 Follow
 11
 3 weeks ago 3w ago
(Original post by DFranklin)
The basic idea looks fine, but some of your arguments are really too vague to really nail this down as a full solution. Words like "clearly" and "should" are always suspect in this kind of thing.
Some thoughts on things you might want to do to clean things up.
If you want to argue that "we can keep iterating", you need to make it very clear that the original conditions are still satisfied. It's kind of trivial, but making a' = b and b' = kba would mean that your argument showed that a' > b' and so we have a more direct parallel with the initial scenario with a > b.
Also, with a "we can keep iterating", you generally want to pay more attention to "when do/should we stop?" In this case you should find your argument for finding b' fails when b = 0. So you can see that this is when the iteration stops and then you have k = a^2, so k is a square.
yes the first point makes complete sense, thats why later in the paragraph I kept having to refer to the 'smaller of the pair' because I used both the letters a and b at points to refer to the smaller of a pair. Thanks
With the second point , my main argument was that rather than rigidly setting constraints for every set of solution pairs to be positive, I set k to be rigidly positive . This would mean that there can never be a transition from (positive not 0, positive not 0) to (positive not 0 , negative), since this would make the denominator ab+1 <=0, making k not positive. But if b' never reached a 0 this WOULD happen since we get decreasing solutions . So for any solutions the sequence passes through 0.
Just to check for no violations we test what happens given reaching 0. When a given b' passes through 0 for the first time for a positive a' , b''= a',a''=0 but his doesn't break our important criteria of k being positive( since ( a'' )( b'' )+1=0+1=1. The fact that a'' is negative isn't too important , it just means we stop before getting to that point (in fact if we kept going we'd just oscillate between (a,0) and (0,a).The important part was that a 0 was reached
The fact that a given b'=0 means that 0=(b^2k)/a meaning that k=b^2
was that a little better?Last edited by chemphys; 3 weeks ago at 11:23. 
 Follow
 12
 3 weeks ago 3w ago
(Original post by chemphys)
was that a little better?
FWIW, while keeping your basic argument I'd probably argue something like this:
Claim: Our sequence must eventually have a pair with
Proof of claim: First note that since this is equivalent to the claim that for some n. Since we start with b_1 > 0 and we have shown , the only way this could fail is if for some n. We show this is impossible. For since (and so is > 0), in this case we would have ; but then since this contradicts our definition of k as a postiive integer.
To prove that k must be a perfect square, all that remains is to note that if then . 
 Follow
 13
 3 weeks ago 3w ago
(Original post by DFranklin)
It's better, and it's probably enough to get the marks, but it's still a bit unclear.
FWIW, while keeping your basic argument I'd probably argue something like this:
Claim: Our sequence must eventually have a pair with
Proof of claim: First note that since this is equivalent to the claim that for some n. Since we start with b_1 > 0 and we have shown , the only way this could fail is if for some n. We show this is impossible. For since (and so is > 0), in this case we would have ; but then since this contradicts our definition of k as a postiive integer.
To prove that k must be a perfect square, all that remains is to note that if then . 
 Follow
 14
 3 weeks ago 3w ago
Basically the method is something called, Vieta Jumping with roots.
Minimal solutions, it is incredibly useful.
Posted from TSR Mobile
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: September 28, 2016
Share this discussion:
Tweet
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.